# If A, B and C are the interior angles of a triangle ABC, show that sin$$B+C\over 2$$ = cos$$A\over 2$$.

## Solution :

Since A, B and C are the interior angles of a triangle ABC

$$\therefore$$   A + B + C = 180

$$\implies$$  $$A\over 2$$ + $$B\over 2$$ + $$C\over 2$$ = 90

$$\implies$$  $$B\over 2$$ + $$C\over 2$$ = 90 – $$A\over 2$$

$$\implies$$  sin($$B+C\over 2$$) = sin(90 – $$A\over 2$$)

$$\implies$$   sin$$B+C\over 2$$ = cos$$A\over 2$$.